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LeetCode 965: Univalued Binary Tree Solution

Master LeetCode problem 965 (Univalued Binary Tree), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To determine if a binary tree is univalued, we need to check if all nodes in the tree have the same value as the root node. We can perform a depth-first search (DFS) traversal of the tree and compare each node's value with the value of the root node. If any node's value is different, we return false. If all nodes have the same value, we return true.

Algorithm:

  1. Initialize a global variable unival to true.
  2. Implement a depth-first search function isUnivalTreeDFS that takes a node and a value as parameters.
  3. In the isUnivalTreeDFS function: a. If the node is null, return true. b. If the node's value is not equal to the given value, set unival to false. c. Recursively call the function on the left and right children of the node.
  4. Call the isUnivalTreeDFS function with the root node's value as the initial value.
  5. Return the value of unival.

Time Complexity: O(n), where n is the number of nodes in the binary tree. Space Complexity: O(h), where h is the height of the binary tree.

:

Solution Code

class Solution {
    boolean unival = true;

    public boolean isUnivalTree(TreeNode root) {
        if (root == null) {
            return true;
        }
        isUnivalTreeDFS(root, root.val);
        return unival;
    }

    private void isUnivalTreeDFS(TreeNode node, int val) {
        if (node == null) {
            return;
        }
        if (node.val != val) {
            unival = false;
        }
        isUnivalTreeDFS(node.left, val);
        isUnivalTreeDFS(node.right, val);
    }
}

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